Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k + 4}{8k - 72} \div \dfrac{k^2 + 5k + 4}{k^2 - 9k} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k + 4}{8k - 72} \times \dfrac{k^2 - 9k}{k^2 + 5k + 4} $ First factor the quadratic. $n = \dfrac{k + 4}{8k - 72} \times \dfrac{k^2 - 9k}{(k + 4)(k + 1)} $ Then factor out any other terms. $n = \dfrac{k + 4}{8(k - 9)} \times \dfrac{k(k - 9)}{(k + 4)(k + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (k + 4) \times k(k - 9) } { 8(k - 9) \times (k + 4)(k + 1) } $ $n = \dfrac{ k(k + 4)(k - 9)}{ 8(k - 9)(k + 4)(k + 1)} $ Notice that $(k - 9)$ and $(k + 4)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ k\cancel{(k + 4)}(k - 9)}{ 8(k - 9)\cancel{(k + 4)}(k + 1)} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $n = \dfrac{ k\cancel{(k + 4)}\cancel{(k - 9)}}{ 8\cancel{(k - 9)}\cancel{(k + 4)}(k + 1)} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $n = \dfrac{k}{8(k + 1)} ; \space k \neq -4 ; \space k \neq 9 $